Question

All the values of '$a$' for which the quadratic expression $a{x}^2+(a-2)x-2$ is negative for exactly two integral values of $x$ may lie in

• A. $(1,3/2)$
• B. $(3,2/2)$
• C. $(1,2)$
• D. $(-1,2)$

Answer 1

The correct option is C $(1,2)$

${ax}^2+(a-2)x-2$

The discriminant is ${(a-2)}^2-4\left(a\right)(-2)=a^2-4a+4+8a=a^2+4a+4={(a+2)}^2$

That's positive for all $a$, so there are always two roots. Now if $a>0$, then the quadratic is negative between those two real roots. Using the quadratic formula for roots :

$\frac{2-a-(a+2)}{2a}<\times <\frac{2-a+a+2}{2a}$

$-1<\times <\frac{2}{a}$

For it exactly $2$ integer values of $x$ are in that interval, the integer must be $0$ and $+1$. That requires $1<2/a$$\le 2$.

Taking reciprocals :

$1>\frac{a}{2}\ge 1/2$ (in equality direction is reversed by reciprocal)

$2>a\ge 1$ (multiply by $2$)

$\therefore$ $aϵ(1,2)$